Loop invariant of insertion sort is: Before $k$th iteration the subarray $A[1:k-1]$ contains the first $k-1$elements from original array in sorted order
Here we have an array $A = [ 1, 3, 5, 7, 9, 11, x, 15, 13]$, after applying insertion sort on $A$, there are exactly two swaps. By seeing the 13 at the end of array we can conclude that there is atleast one swap happens in the last iteration and another thing here is the subarray $A[0: 5]$ is already sorted.
Case 1: all the two swaps happens in the last iteration$(8th \text{ iteration } )$. Then it means before the $8\text{th}$ iteration the array $A[0:7]$ is sorted then implies $x \leq 15$ and since last two swaps must happen in last iteration, one swap happens with $15$ and $13$ and other swap must be $13$ and $x$, which means $x > 13 $. so there are two conditions $x > 13 \text{ and }x \leq 15$, the possible $x's $ are $14, 15$.
Case 2: Exactly one swap happens in the last iteration$\implies x \leq 13$, the remainging swap must happen in $6\text{th}$ iteration, because if there is swap in $\text{7th}$ iteration then implies $x >15$, which is a contradiction. So we need exactly one swap in $\text{6th}$ iteration $\implies x < 11 \text{ and } x \geq 9 \implies $ the possible $x's$ are $9, 10 $.
In the given options, the answers are $A , C$
